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I solved for the range as follows:

Setting $f(x)=\frac {2x^2+x-3}{x^2+4x-5} =y$, I rearranged it to get a quadratic in x. $$(y-2)x^{2}+ (4y-1)x +(3-5y)=0$$
Next, using $\Delta \ge 0$, I got $$(4y-1)^2-4(y-2)(3-5y) \ge 0$$ Which boiled down to
$$(6y-5)^2 \ge 0$$ And this gave me $$ y \in R$$ Next, the value of $x$ for which $y=2$ is $x=1$, for which the function isn't defined, so that gives me $$y \in R - \{2\}$$ However, the solution is $$y \in R- \{\frac{5}{6}, 2\}$$ I understand that the original function is identical to $$g(x) =\frac{2x+3}{x+5} \forall x \in R - \{1\}$$
and that $\frac{5}{6} =g(1)$, but in the original function $f(x)$ I found $y=2$ corresponded to $x=1$ and therefore excluded it, but if $x= 1$ also corresponds to $y=\frac{5}{6}$ then wouldn't this $not$ be a function, as $x=1$ would then be associated with two different $y$ values?

I've read the answers in Why D≥0 while finding the range of rational functions and Finding the range of $y =\frac{x^2+2x+4}{2x^2+4x+9}$ (and $y=\frac{\text{quadratic}}{\text{quadratic}}$ in general) but I'm still unsure of how to apply the information from those to figure out what values of $y$ need to be excluded when dealing with such questions.

While I am familiar with derivatives and limits to a certain degree, we were assumed to $\underline {not}$ know calculus when we were taught this and solved such problems.

I apologise if there are any issues with formatting, this is my first time using Latex.

Kenobi
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    "In the original function $f(x)$ I found $y=2$ corresponded to $x=1$". Shouldn't that be $x=2$ corresponds to $y=1$...it makes a difference! – paw88789 Jul 14 '22 at 14:00
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    The first step is to write this as partial fractions $f(x)=2-\frac{7}{x+5}$ and to look at the usual hyperbolic decay of the rational polynomial. – R. J. Mathar Jul 14 '22 at 14:08
  • @paw88789 I did mean $y=2$ and $x=1$, because when I plug $y=2$ I'm left with $7x-7=0$ which gives me $x=1$.
    What I'd learnt was that I should plug in the value of $y$ that would cause the leading coefficient in the quadratic to vanish, and check if the resulting value of $x$ is still in the domain of the original function, and then to include/exclude that $y$ from the range.     – Kenobi Jul 14 '22 at 14:58
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    @Kenobi I see that works for the equation after rearrangement, but not in the original function (since $f(1)$ is not defined). In fact, in the rearranged equation ($(y-2)x^2+(4y-1)x+(3-5y)=0$), you can plug in any value for $y$ and get $x=1$ as a corresponding value for $x$. Note that when $x=1$ the rearranged equation simplifies to $0=0$. – paw88789 Jul 14 '22 at 15:11
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    You can write $\mathbb{R}$ in math jax with \mathbb{R}. – Etemon Jul 14 '22 at 16:26

3 Answers3

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Your quadratic equation has $x=1$ as a solution, for any $y$. Indeed, for $y\neq2$ the quadratic formula gives

$$x=\frac{-(4y-1)\pm(6y-5)}{2(y-2)}$$ $$x=\frac{-10y+6}{2y-4}=\frac{-(5y-3)}{y-2}\qquad\text{or}\qquad x=\frac{2y-4}{2y-4}=1$$

and for $y=2$ your equation reduces to $7x-7=0$.

The original function was

$$y=\frac{(x-1)(2x+3)}{(x-1)(x+5)}$$

and you rearranged this to get

$$(x-1)(x+5)y-(x-1)(2x+3)=0$$

which always has $x=1$ as a solution.

Essentially, you added a vertical line to the graph of $g$, so the resulting graph doesn't represent a function.

You should have just worked with $g$ from the beginning, and then excluded any values of $y$ which would have given $x=1$ or $x=-5$.

mr_e_man
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  • So is the rearranging method incorrect? Because that was what I learnt at school, unless I've misinterpreted what was taught. – Kenobi Jul 14 '22 at 14:51
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    You can rearrange things like that, but you still have to exclude values of $y$ which would give values of $x$ not in the original domain. – mr_e_man Jul 14 '22 at 14:56
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    (in this case, $y=5/6$) – mr_e_man Jul 14 '22 at 14:59
  • How would I know to test for $y = \frac{5}{6}$ in the rearranged quadratic $$(x-1)(x+5)y - (x-1)(2x+3)=0 ?$$ – Kenobi Jul 14 '22 at 15:03
  • I was taught to turn the thing into a quadratic and then solve $\Delta \ge 0$ and then as a step-2 I should plug the value of $y$ which causes the leading coefficient to vanish and check if the resulting $x$ value is in the domain. Doing that led me to exclude $y=2$ but I'm unsure how I'd be able to know to test for values like $y= \frac{5}{6}$ – Kenobi Jul 14 '22 at 15:06
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    Start with the equation $x=-(5y-3)/(y-2)$. Setting $x=1$ and solving for $y$, you'll get $y=5/6$. Setting $x=-5$, you'll find there is no solution for $y$. – mr_e_man Jul 14 '22 at 15:25
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    Just keep in mind the definition of the range: For any $y$ in the range of $f$, there exists at least one $x$ in the domain of $f$, such that $f(x)=y$. You should be able to figure everything out, based on that. – mr_e_man Jul 14 '22 at 15:25
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Alternative approach, along the lines of R. J. Mathar in the comments.

For all $x\in\mathbb{R}\setminus\{1,-5\},$

$$ \frac{2x^2 + x -3}{x^2 + 4x - 5} \equiv \frac{2(x^2 + 4x - 5) - 7x + 7}{x^2 + 4x - 5} \equiv 2 - \frac{7x-7}{x^2 + 4x - 5} $$

$$ \equiv 2 - \frac{7(x-1)}{(x+5)(x-1)} \equiv 2 - \frac{7}{(x+5)}. $$

Define $g:x\to 2 - \frac{7}{(x+5)}\ $ and $\ f:x\to\frac{2x^2 + x -3}{x^2 + 4x - 5}.$

We showed above that $f$ and $g$ are equivalent for all $x\in\mathbb{R}\setminus\{1,-5\}.$

The function $g:x\to 2 - \frac{7}{(x+5)}$ has domain $\mathbb{R}\setminus\{-5\}$ and range $\mathbb{R}\setminus\{2\}.$

But the function $ f:x\to\frac{2x^2 + x -3}{x^2 + 4x - 5}$ has domain $\mathbb{R}\setminus\{1,-5\}.$ Therefore the range of $f$ is the range of $g$ with the point $ g(1)$ (if it exists) removed.

Adam Rubinson
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  • So, my general approach should be to find an 'easier' function which is identical to the given function except for a few points, and then consider and exclude said points from the range? – Kenobi Jul 14 '22 at 15:11
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    Well that was my approach. You can use whichever approach you want. – Adam Rubinson Jul 14 '22 at 15:35
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As you have already understood x -1 is a factor of both numerator and denominator of f(x) . Therefore 1 can not be in the domain of the function f . Actually what is happening here is f(x) is indeterminate when x = 1 and it can take any value. Therefore if you define the domain of f as ℝ - {1} that issue doesn't arise and you can consider f as a function.

Janaka Rodrigo
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